On which interval(s) is the function y = x^3 + 3/2x^2 b) (-inf, -2); (1, inf) c) (-inf, inf) d) (-2, 1) e)...

y= x^3 + (3/2)x^2 - 6x +
27

First we will find the first
derivative.

==> y' = 3x^2 + 3x -
6

Now we will determine the
zeros.

==> 3x^2 + 3x - 6 =
0

Divide by 3:

==>
x^2 + x -2 = 0

==>  (x+2)(x-1) =
0

==> x= -2 and x=
1

Then the function changes direction at the points x=-2
and x= 1

Then we have the following increasing or
decreasing intervals.

==> ( -inf, -2) ( -2, 1) and (
1, inf)

==> The function increasing when y' >
0

==> (x+2)(x-1) >
0

==> x> -2 and x >
1

==> x >
1

OR

==> x < -2
and x < 1

==> x <
-2

Then the function is increasing when x<-2 and x
> 1

==> x = (-inf, -2) U  (1,
inf)

Then the answer is: b) ( -inf,-2) ; ( 1,
inf).